what is an easy way to find the asymptotes of the graph $r = \frac1{1+2\cos \theta}?$ i would like to know an easy way to find the asymptotes of the $$r = \frac1{1+2\cos \theta}.$$   this is for a precalculus audience, so it will be nice if calculus can be avoided altogether.
i know that the asymptotes are parallel to the lines $\theta = \pm \pi/3.$  we also know that the $x$-intercepts are $1/3, 1.$
my audience don't know how to use the symmetry, most of them won't even know that the graph is a hyperbola.  can we find the answer without using the symmetry of the graph. 
 A: Change basis to $\,\vec u_{2\pi/3}$ (the unit vector with polar angle $\frac{2\pi}3$) and the orthogonal vector $\vec v_{2\pi/3}$. An asymptote with polar angle $\frac{2\pi}3$  then has equation $Y=r\sin\bigl(\theta-\frac{2\pi}3\bigr)=l$. Hence you have to determine:
$$\lim_{\theta\to\tfrac{2\pi}3}r(\theta)\sin\Bigl(\theta-\frac{2\pi}3\Bigr). $$
With some trigonometry ($\cos p-\cos q= \dots$), you should find a limit equal to $\frac 1{\sqrt 3}$.
A: here is what i came up with. we know that the graph of $r = \frac{1}{1+2\cos \theta}$ as a function of $\theta$ has vertical asymptotes at $\theta = 2\pi/3, 4\pi/3.$ 
we will show that for $\theta$ near $2\pi /3$ the polar graph is the graph of the slant asymptote of the hyperbola that cuts the $x$-axis at $\frac23.$ 
let $ \theta = 2\pi/3 + h.$ then 
$$\begin{align}
r &= \frac1{1 + 2\cos(2\pi/3+h)}\\
&=\frac1{1+2\left(\cos(2\pi/3)\cos h - \sin(2\pi/3)\sin h \right)} \\
&=-\frac1{\sqrt3\sin h + \cdots} \\
&=-\frac1{\sqrt 3 \sin (t-2\pi/3) + \cdots}\end{align}$$
therefore one of the asymptotes is $$r =-\frac1{\sqrt3\sin (t-2\pi/3)},\, r(0) = \frac23 $$ as claimed.
in the same way we find that the second asymptote is the line 
$$r = \frac1{\sqrt3 \sin(t - 4\pi/3)}, \, r(0) = \frac23$$
and we see that the two asymptotes meet at the $x$-intercept $(\frac23, 0).$
